Text Chapter 4 Examples (1)

Example 4.1, p.87. Let X represent the uncertainty in the number of delivered source instructions (DSI) of a new software applica Suppose the uncertainty is expressed as the trapezoidal density function seen below. Determine the following A. 𝐸(𝑋) . B. Med(𝑋) . C. 𝑃(𝑋≤𝐸(𝑋)+𝜎_𝑋) . PDF, CDF, Mean, and Variance:

A. a 25,000 parameters of distribution m1 28,000 m2 35,000 b 37,500 n1 3,943,750,000 numerator 1st term n2 2,109,000,000 numerator 2nd term d 58,500 denominator E(X) 31,363.2479 31,363 n1 1.9076563E+14 numerator 1st term n2 7.4677E+13 numerator 2nd term d 117000 denominator [E(X)]^2 983,653,317 Var(X) 8557153.55395 need this for part C Sigma(X) 2925.2612796 2,925 need this for part C B. First we compute the area of the triangle on the LHS. base 3,000 height 0.0001025641 =2/((35000+37500)-(25000+28000)) Area 0.1538 Now we compute the area of the triangle on the RHS. base 2,500 height 0.0001025641 Area 0.1282 P(25,000 <= x <= 28,000) 0.1538 P(35,000 <= x <= 37,500) 0.1282 P(28,000 <= x <= 35,000) 0.7179 P(25,000 <= x <= 35,000) 0.8718 * The text has 34/39 which equals 0.8718. Based on the above, the median includes some portion of the rectangle of the trapezoid. So, we use the CDF with m1 <= x <= m2. a 25,000 m1 28,000 m2 35,000

b 37,500 CDF term1 5.1282051E-05 CDF term2 2x - 53,000 0.5 Equation term1 x (2x - 53,000) = 0.50 Med(X) 31,375 C. We need to determine P(X <= E(X) + Sigma(X)) E(X) + Sigma(X) 34,288 We can see that the E(X) + Sigma(X) is in the rectangular region of the trapezoidal distribution. a 25,000 m1 28,000 m2 35,000 b 37,500 term1 5.1282051E-05 term2 15576 P(X <= 34,288) 0.7988

ation. g: Example 4.2, p.91. If 𝑋 has a uniform distribution, show that 𝑀𝑒𝑑(𝑋)=𝐸(𝑋). The PDF for a uniform distribution is: ?_𝑋 (𝑥)=1/(𝑏−𝑎), 𝑖? 𝑎≤𝑥≤𝑏 Recall that we integrate the PDF to get the CDF (we differentiate the CDF to get the PDF): ?_𝑋 (𝑥)=1/(𝑏−𝑎) ∫1_𝑎^𝑥▒𝑑𝑢=1/(𝑏−𝑎) ├ 𝑢┤|_𝑎^𝑥 ?_𝑋 (𝑥)=1/(𝑏−𝑎) (𝑥−𝑎), 𝑖? 𝑎≤𝑥≤𝑏 The 𝑀𝑒𝑑(𝑋)=0.50. So, we take the CDF and set it equal to 0.50 and solve for 𝑥 . 1/(𝑏−𝑎) (𝑥−𝑎)=0.50 𝑥=1/2 (𝑏−𝑎)+𝑎, 𝑥=1/2 𝑏−1/2 𝑎+𝑎=1/2 𝑏+1/2 𝑎=(𝑏+𝑎)/2 Thus: 𝑀𝑒𝑑(𝑋)=(𝑏+𝑎)/2 Finally, we have that the E(𝑋) is: 𝐸(𝑋)=∫1_(−∞)^∞▒ 𝑥∙? 〖 _𝑋 (𝑥)𝑑𝑥=∫1_𝑎^𝑏▒ 𝑥 〖 /(𝑏−𝑎) 𝑑𝑥=1/(2(𝑏−𝑎))∙├ 𝑥^2 ┤|_𝑎^𝑏= 〗 (𝑏^2−𝑎^2)/(2(𝑏−𝑎))=((𝑏+𝑎)(𝑏−𝑎))/(2(𝑏−𝑎))=(𝑏+𝑎)/2 So, for a uniform distribution Med(𝑋)=E(𝑋)=(𝑏+𝑎)/2.